Question: $\lim_{x\to 3}\dfrac{\sqrt{2x-5}-1}{x-3}=$
Substituting $x=3$ into $\dfrac{\sqrt{2x-5}-1}{x-3}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since we have a rational expression with a square root on our hands, let's try to re-write it using the method of rationalization. $\begin{aligned} &\phantom{=}\dfrac{\sqrt{2x-5}-1}{x-3} \\\\ &=\dfrac{\sqrt{2x-5}-1}{x-3}\cdot\dfrac{\sqrt{2x-5}+1}{\sqrt{2x-5}+1} \gray{\text{Rationalize the numerator}} \\\\ &=\dfrac{(2x-5)-1^2}{(x-3)(\sqrt{2x-5}+1)} \\\\ &=\dfrac{2\cancel{(x-3)}}{\cancel{(x-3)}(\sqrt{2x-5}+1)} \gray{\text{Cancel out common factors}} \\\\ &=\dfrac{2}{\sqrt{2x-5}+1} \text{, for }x\neq 3 \end{aligned}$ This means that the two expressions have the same value for all $x$ -values (in their domains) except for $3$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{\sqrt{2x-5}-1}{x-3}=\dfrac{2}{\sqrt{2x-5}+1}$ for all $x$ -values in the interval $(2.5,3.5)$ except for $x=3$. Therefore, $\lim_{x\to 3}\dfrac{\sqrt{2x-5}-1}{x-3}=\lim_{x\to 3}\dfrac{2}{\sqrt{2x-5}+1}=1$. (The last limit was found using direct substitution.) [I want to see how this looks graphically!] In conclusion, $\lim_{x\to 3}\dfrac{\sqrt{2x-5}-1}{x-3}=1$.